Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/gen-9)

The rewrite relation of the following TRS is considered.

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

• The 1^st component contains the pair

  f#(c(x,y,z))

  →

  f#(b(y,z))

  (6)

  1.1.1 Reduction Pair Processor with Usable Rules

  Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

  [f#(x1)]	=	0 0 + 0 1 0 0 · x1
  [c(x1, x2, x3)]	=	0 1 + 1 0 1 0 · x1 + 1 0 1 0 · x2 + 0 0 1 1 · x3
  [b(x1, x2)]	=	1 0 + 1 0 1 0 · x1 + 1 0 1 0 · x2
  [a]	=	1 1
  [f(x1)]	=	0 0 + 1 1 1 0 · x1

  together with the usable rules

  b(z,b(c(a,y,a),f(f(x))))	→	c(c(y,a,z),z,x)	(3)
  c(c(z,y,a),a,a)	→	b(z,y)	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the pair

  f#(c(x,y,z))

  →

  f#(b(y,z))

  (6)

  could be deleted.

  1.1.1.1 P is empty

  There are no pairs anymore.

• The 2^nd component contains the pair

  b#(z,b(c(a,y,a),f(f(x))))	→	c#(c(y,a,z),z,x)	(8)
  c#(c(z,y,a),a,a)	→	b#(z,y)	(4)
  b#(z,b(c(a,y,a),f(f(x))))	→	c#(y,a,z)	(9)

  1.1.2 Reduction Pair Processor with Usable Rules

  Using the Knuth Bendix order with w0 = 4 and the following precedence and weight functions

  prec(b)	=	1		weight(b)	=	11
  prec(c)	=	2		weight(c)	=	8
  prec(a)	=	0		weight(a)	=	4
  prec(f)	=	3		weight(f)	=	7

  in combination with the following argument filter

  π(b#)	=	2
  π(b)	=	[1,2]
  π(c)	=	[1,2]
  π(a)	=	[]
  π(f)	=	[]
  π(c#)	=	1

  together with the usable rules

  b(z,b(c(a,y,a),f(f(x))))	→	c(c(y,a,z),z,x)	(3)
  c(c(z,y,a),a,a)	→	b(z,y)	(1)

  (w.r.t. the implicit argument filter of the reduction pair), the pairs

  b#(z,b(c(a,y,a),f(f(x))))	→	c#(c(y,a,z),z,x)	(8)
  c#(c(z,y,a),a,a)	→	b#(z,y)	(4)
  b#(z,b(c(a,y,a),f(f(x))))	→	c#(y,a,z)	(9)

  could be deleted.

  1.1.2.1 P is empty

  There are no pairs anymore.